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Pavlova-9 [17]
2 years ago
12

Please help! Will mark brainliest <3. (Click on the photo too see the question!) you need to select all angle pairs that are

vertical angles.

Mathematics
2 answers:
wariber [46]2 years ago
7 0

Answer:

B and D

Step-by-step explanation:

vertical angles are ones that are formed by intersecting lines, and they are also equal to each other and opposite of each other.

its not A because they are not equal to each other.

B is correct because they are formed by intersecting line and equal to each other

C is incorrect since theres some random line that doesnt completely pass through on either side

D is correct

sattari [20]2 years ago
4 0

Answer:

☐ \displaystyle \textcolor{blue}{A.}\:\angle{S}\:and\: \angle{J}

☑ \displaystyle \textcolor{blue}{B.}\:\angle{G}\:and\: \angle{M}

☐ \displaystyle \textcolor{blue}{C.}\:\angle{H}\:and\: \angle{P}

☑ \displaystyle \textcolor{blue}{D.}\:\angle{R}\:and\: \angle{U}

Explanation:

Accourding to the Vertical Angles Theorem, vertical angles are considered congruent. So, we need to make sure we find all pairs of angles that <em>reflect</em> each other like a mirror [NO LINEAR INTERFERENSES], and those would be the above checked checkboxes.

I am joyous to assist you at any time.

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Please answer. I will mark Brainliest
den301095 [7]
Statement: 1) Triangle ABC
2) 2x 3)x=52

Reason: I’m not sure but probably addition
4 0
2 years ago
Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?
Bogdan [553]

Step-by-step explanation:

\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}

\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}

\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}

\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}

\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}

\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}

Taking sin²θ common in both numerator & denominator, We get :

\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}

\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}

\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}

\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}

\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}

\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}

\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}

\mathsf{\implies \dfrac{5 - 1}{5 + 1}}

\mathsf{\implies \dfrac{4}{6}}

\mathsf{\implies \dfrac{2}{3}}

<u>Hence</u><u>,</u><u> option</u><u> </u><u>(</u><u>a)</u><u> </u><u>2</u><u>/</u><u>3</u><u> </u><u>is </u><u>your</u><u> </u><u>correct</u><u> </u><u>answer</u><u>.</u>

3 0
2 years ago
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